Download Permutation and Combination Problems with solutions pdf. Today, I am going to share techniques to solve permutation and combination questions. This chapter talk about selection and arrangement of things which could be any numbers, persons,letters,alphabets,colors etc.The basic difference between permutation and combination is of order.
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Download Permutation and Combination Problems with solutions pdf (link given below post)
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1. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? |
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| A. 24400 | B. 21300 |
| C. 210 | D. 25200 |
Answer With Explanation :
Answer: Option D
Explanation:
Number of ways of selecting 3 consonants from 7
= 7C3
Number of ways of selecting 2 vowels from 4
= 4C2
Number of ways of selecting 3 consonants from 7 and 2 vowels from 4
= 7C3 × 4C2
=(7×6×53×2×1)×(4×32×1)=210=(7×6×53×2×1)×(4×32×1)=210
It means we can have 210 groups where each group contains total 5 letters (3 consonants and 2 vowels).
Number of ways of arranging 5 letters among themselves
=5!=5×4×3×2×1=120=5!=5×4×3×2×1=120
Hence, required number of ways
=210×120=25200=210×120=25200
| 2. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there? | |
| A. 159 | B. 209 |
| C. 201 | D. 212 |
Answer With Explanation :
Answer: Option B
Explanation:
In a group of 6 boys and 4 girls, four children are to be selected such that at least one boy should be there.
Hence we have 4 options as given below
We can select 4 boys …(option 1)
Number of ways to this = 6C4
We can select 3 boys and 1 girl …(option 2)
Number of ways to this = 6C3 × 4C1
We can select 2 boys and 2 girls …(option 3)
Number of ways to this = 6C2 × 4C2
We can select 1 boy and 3 girls …(option 4)
Number of ways to this = 6C1 × 4C3
Total number of ways
= 6C4 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C3
= 6C2 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C1[∵ nCr = nC(n-r)]
=6×52×1+6×5×43×2×1×4=6×52×1+6×5×43×2×1×4 +6×52×1×4×32×1+6×4+6×52×1×4×32×1+6×4
=15+80+90+24=209=15+80+90+24=209
| 3. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there in the committee. In how many ways can it be done? | |
| A. 624 | B. 702 |
| C. 756 | D. 812 |
Answer With Explanation
Answer: Option C
Explanation:
From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
Hence we have the following 3 options.
We can select 5 men …(option 1)
Number of ways to do this = 7C5
We can select 4 men and 1 woman …(option 2)
Number of ways to do this = 7C4 × 6C1
We can select 3 men and 2 women …(option 3)
Number of ways to do this = 7C3 × 6C2
Total number of ways
= 7C5 + (7C4 × 6C1) + (7C3 × 6C2)
= 7C2 + (7C3 × 6C1) + (7C3 × 6C2)[∵ nCr = nC(n – r) ]
=7×62×1+7×6×53×2×1×6=7×62×1+7×6×53×2×1×6 +7×6×53×2×1×6×52×1+7×6×53×2×1×6×52×1
=21+210+525=756=21+210+525=756
Permutation and Combination Problems with solutions pdf (link given below )
| 4. In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together? | |
| A. 610 | B. 720 |
| C. 825 | D. 920 |
Answer With Explanation
Answer: Option B
Explanation:
The word ‘OPTICAL’ has 7 letters. It has the vowels ‘O’,’I’,’A’ in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA).
Hence we can assume total letters as 5 and all these letters are different.
Number of ways to arrange these letters
=5!=5×4×3×2×1=120=5!=5×4×3×2×1=120
All the 3 vowels (OIA) are different
Number of ways to arrange these vowels among themselves
=3!=3×2×1=6=3!=3×2×1=6
Hence, required number of ways
=120×6=720=120×6=720
| 5. In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together? | |
| A. 47200 | B. 48000 |
| C. 42000 | D. 50400 |
Answer With Explanation
Answer: Option D
Explanation:
The word ‘CORPORATION’ has 11 letters. It has the vowels ‘O’,’O’,’A’,’I’,’O’ in it and these 5 vowels should always come together. Hence these 5 vowels can be grouped and considered as a single letter. that is, CRPRTN(OOAIO).
Hence we can assume total letters as 7. But in these 7 letters, ‘R’ occurs 2 times and rest of the letters are different.
Number of ways to arrange these letters
=7!2!=7×6×5×4×3×2×12×1=2520=7!2!=7×6×5×4×3×2×12×1=2520
In the 5 vowels (OOAIO), ‘O’ occurs 3 and rest of the vowels are different.
Number of ways to arrange these vowels among themselves =5!3!=5×4×3×2×13×2×1=20=5!3!=5×4×3×2×13×2×1=20
Hence, required number of ways
=2520×20=50400=2520×20=50400
6 – How many 3 digit number can be formed with the digits 5, 6, 2, 3, 7 and 9 which are
divisible by 5 and none of its digit is repeated?
a) 12
b) 16
c) 20
d) 24
e) None of these
Answer & Explanation
Answer – c) 20
Explanation :
_ _ 5
first two places can be filled in 5 and 4 ways respectively so, total number of 3 digit number =
5*4*1 = 20
7- In how many different ways can the letter of the word ELEPHANT be arranged so that
vowels always occur together?
a) 2060
b) 2160
c) 2260
d) 2360
e) None of these
Answer & Explanation
Answer – b) 2160
Explanation :
Vowels = E, E and A. They can be arranged in 3!/2! Ways
so total ways = 6!*(3!/2!) = 2160
Permutation and Combination Problems with solutions pdf (link given below post)
8- There are 4 bananas, 7 apples and 6 mangoes in a fruit basket. In how many ways can a
person make a selection of fruits from the basket.
a) 269
b) 280 c) 279
d) 256
e) None of these
Answer & Explanation
Answer – c) 279
Explanation :
Zero or more bananas can be selected in 4 + 1 = 5 ways (0 orange, 1 orange, 2 orange, 3
orange and 4 orange)
similarly apples can be selected in 7 +1 = 8 ways
and mangoes in 6 +1 = 7 ways
so total number of ways = 5*8*7 = 280
but we included a case of 0 orange, 0 apple and 0 mangoes, so we have to subtract this, so 280
– 1 = 279 ways
9- There are 15 points in a plane out of which 6 are collinear. Find the number of lines that
can be formed from 15 points.
a) 105
b) 90
c) 91
d) 95
e) None of these
Answer & Explanation
Answer – c) 91
Explanation :
From 15 points number of lines formed = 15c2
6 points are collinear, number of lines formed by these = 6c2
So total lines = 15c2 – 6c2 + 1 = 91